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skyhimself75

  • 3 years ago

what is the solution of 5|x+1|+6=21

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  1. amistre64
    • 3 years ago
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    how would we start it?

  2. Alexis1994_13
    • 3 years ago
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    5|χ+1|=15 => |x+1| = 3.

  3. amistre64
    • 3 years ago
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    im sure as long as you just get an answer; you will continue to type in the same kind of problem over and over again :)

  4. Llort
    • 3 years ago
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    assuming x is real: \[5 \sqrt((x+1)^2)+6 = 21\] otherwise: |x+1|=3

  5. rauch
    • 3 years ago
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    2

  6. Llort
    • 3 years ago
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    don't just give it to him.

  7. skyhimself75
    • 3 years ago
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    i don't know how to start it absolute value i just don't understand it

  8. amistre64
    • 3 years ago
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    the absolute value is the last thing to work on; how do you get that absolute value all by itself?

  9. rauch
    • 3 years ago
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    5(x+1)+6=21 (5x+5)+6=21 5x+6=16 5x=10 x=2

  10. Llort
    • 3 years ago
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    looking at this question, what is the other solution for x if it was not an absolute value?

  11. skyhimself75
    • 3 years ago
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    i'm seriously terrible in math i really don't know:[

  12. amistre64
    • 3 years ago
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    the term "absolute value" means that there is a quantifiable value; something that can be measured. length, height, area, volume, distance are all examples of abslute values

  13. jgeorge123
    • 3 years ago
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    dont jus give him the answer he is probably usin this site for a summer school!!!

  14. amistre64
    • 3 years ago
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    in other words; an absolute value doesnt care about what direction the sign takes us in; it only cares about a quantifiable measure

  15. Llort
    • 3 years ago
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    |a| is always positive, you discount the negative answer.

  16. polpak
    • 3 years ago
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    sky, you isolate the absolute value just like you would isolate a variable. Llort, no you do not 'discount' the negative answer. It is a valid solution.

  17. amistre64
    • 3 years ago
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    |-12| has the same absolute value as |12|

  18. amistre64
    • 3 years ago
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    as a result; if we are determining any value for inside of the absolute value notation; we have to account for the positive and the negative results that occur within it

  19. polpak
    • 3 years ago
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    So if you have \[5|x+1| + 6 = 21\] You want to isolate the term with the absolute value in it first. You do this by subtracting 6 from both sides of the equation.

  20. polpak
    • 3 years ago
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    What do you get when you subtract the 6?

  21. skyhimself75
    • 3 years ago
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    -2?

  22. polpak
    • 3 years ago
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    No, \(5|x+1| + 6 - 6 = 21 -6\) \[\implies 5|x+1| + 0 = 15\]\[\implies 5|x+1| = 15\]

  23. polpak
    • 3 years ago
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    So now lets say that |x+1| is something, lets call it k.. So we have \(5k = 15 \)\[\implies \frac{1}{5}\cdot5k = \frac{1}{5}15\]\[\implies k = 3\] And since we said k was |x+1| we have \[|x+1| = 3\]So what values can x have and still make this equation true?

  24. Llort
    • 3 years ago
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    I am assuming the other answer is x=-4

  25. polpak
    • 3 years ago
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    x =-4 is one solution yes, because |-4 + 1 | = |-3| = 3

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